Question

If $C_0,C_1,C_2,..............C_n$ are binomial coefficients then $\frac{1}{n!0!}+\frac{1}{(n-1)!1!}+\frac{1}{(n-2)!2!}+....+\frac{1}{0!n!}$ is equal to

• A. $2^n$
• B. $\frac{2^{n-1}}{n!}$
• C. $\frac{2^n}{n!}$
• D. none of these

Answer 1

The correct option is C $\frac{2^n}{n!}$

Multiplying and dividing the entire expression by n!, we get
$\frac{1}{n!}[\frac{n!}{0!.n!}+\frac{n!}{(n-1)!.1!}+\frac{n!}{(n-2)!.2!}+...\frac{n!}{0!.n!}]$
$=\frac{1}{n!}[{}^n{C}_0+{}^n{C}_1+{}^n{C}_2+...{}^n{C}_n]$
$=\frac{1}{n!}(1+x{)}_{x=1}^n$
$=\frac{2^n}{n!}$.